Question

Hi, can you help to find (all the roots/zeros), please!!!

Answer 1

Given a quadratic equation;

`ax^2+bx+c=0`

The roots x of the equation is;

`x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Given;

`x^2-2x-12=0`
`a=1,b=-2,c=-12`

The values of x are;

`\begin{gathered} x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1)(-12)}}{2(1)} \\ x=\frac{2\pm\sqrt[]{4+48}}{2(1)} \\ x=\frac{2\pm\sqrt[]{52}}{2} \\ x=\frac{2\pm2\sqrt[]{13}}{2} \\ x=1\pm\sqrt[]{13} \end{gathered}`

Then, we can separate the solution;

`x=1+\sqrt[]{13},x=1-\sqrt[]{13}`

Hence, the roots of the quadratic equation are;

`\begin{gathered} x=1+\sqrt[]{13} \\ x=1-\sqrt[]{13} \end{gathered}`